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# how is this possible.2+2=5?

I herd from someone that this is the right anewer to this proble is this true.

### 18 Answers

- csucdartgirlLv 71 decade agoFavorite Answer
This phrase is used in business, not in Math. It is used to illustrate good mergers. When a peanut butter company and a jelly company combine you could get a 2 + 2 = 5 situtation. The combination is logical and useful to each other.

But if the peanut butter company and a bullet manufacturer combined it would only be 2 + 2 = 4. There is no mutual benefit.

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- 1 decade ago
Congruum Problem

Date: 04/04/2002 at 18:06:35

From: Dr. Allan

Subject: Congruum problem

In researching suitable problems for my students I have found a

reference to Fibonacci and his congruum problem. Something has me

stumped; I hope one or more of you can tell me what I am doing wrong:

Quote from the article at the MacTutor History of Mathematics archive

at St. Andrews University:

http://www-history.mcs.st-and.ac.uk/history/Mathem...

"[Fibonacci] defined the concept of a congruum, a number of the form

ab(a + b)(a - b), if a + b is even, and 4 times this if a + b is odd

where a and b are integers. Fibonacci proved that a congruum must be

divisible by 24 and he also showed that for x,c such that x^2 + c and

x^2 - c are both squares, then c is a congruum. He also proved that a

square cannot be a congruum."

With x=15 and c=216, we get the two squares 441 and 9, meaning that

216 should be a congruum. Thus we should be able to find numbers a

and b such that

216 = ab(a+b)(a-b) if a+b is even

or

54 = ab(a+b)(a-b) if a+b is odd

Now, b must be smaller than a and d:= ab(a+b)(a-b) is increasing in

both a and b.

Let's look at the possible scenarios.

Say a+b is odd. We want d to equal 54. Our triples (b,a,d) give us

(1,2,6) (1,4,60); thus b=1 is not possible

(2,3,30) (2,5,210); thus b=2 is not possible

(3,4,84); thus b = 3 and b > 3 is not possible (since d is

increasing); thus a+b odd is not possible

Say a+b is even. We want d to equal 216. Our triples (b,a,d) give us

(1,5,120) (1,7,336); thus b=1 is not possible

(2,4,96) (2,6,3849; thus b=2 is not possible

(3,5,360); thus b = 3 and b > 3 is not possible (since d is

increasing); thus a+b even is not possible

So, because 15^2 + 216 = 21^2 and 15^2 - 216 = 3^2 we should have

that 216 is a congruum, but we cannot put it in the form ab(a+b)(a-b).

What is wrong with this argument?

Sincerely,

Dr. Allan

--------------------------------------------------------------------------------

Date: 04/05/2002 at 16:04:44

From: Dr. Paul

Subject: Congruum problem

I went to the library this morning and picked up a copy of Ore's

_Number Theory and its History_. The solution to the congruum problem

is listed on pages 188-193 and I found it very interesting reading.

What follows is essentially Ore's solution with my commentary added

where I felt Ore's details were (probably intentionally) lacking

sufficient explanation:

We want to find a number x such that simultaneously:

x^2 + h = a^2, x^2 - h = b^2 (equation 8-16)

and determine for which h rational solutions x can exist. We shall

first determine the solutions in integers, and this depends, as we

shall see again, on the Pythagorean triangle. When the second equation

in (8-16) above is subtracted from the first, one has:

2*h = (a^2 - b^2) = (a-b)*(a+b) (8-17)

Since the left-hand side is even, a and b must both be odd or both be

even. Therefore, a - b is even. Hence

a - b = 2*k

and k must be a divisor of h since according to (8-17) we have:

2*h = 2*k*(a+b)

or

h = k*(a+b)

which implies k divides h.

It follows that

(a + b) = h/k

and by adding and subtracting the last two equations, one finds

a = h/(2*k) + k, b = h/(2*k) - k

When these two expressions are substituted into the original equations

(8-16), we obtain:

x^2 + h = [h/(2*k) + k]^2 = [h/(2*k)]^2 + h + k^2

and

x^2 - h = [h/(2*k) - k]^2 = [h/(2*k)]^2 - h + k^2

Adding these equations and removing a common factor of two gives:

x^2 = [h/(2*k)]^2 + k^2

In our case of interest, we have:

x = 15

h = 216

a = 21

b = 3

k = 9

Therefore, the three numbers

x, h/(2*k), k

form a Pythagorean triangle. Determine the lesser of h/(2*k) and k (in

this case k = 9 < 12 = 216/18 = h/(2*k)) so that we can write:

x = t*(m^2 + n^2)

k = t*(m^2 - n^2) (**)

h/(2*k) = 2*m*n*t

where t is some integer and the expressions in m and n define a

primitive solution of the triangle.

In actuality, we have:

x = 15

k = 9

h = 216

so clearly, t = 3 and this forces m = 2, n = 1

As an aside, if h/(2*k) > k then we pick

h/(2*k) = t*(m^2 - n^2)

k = 2*m*n*t

Now we take the product of the last two expressions in (**) above to

obtain as the general solution to (8-16)

x = t*(m^2 + n^2), h = 4*m*n*(m^2 - n^2)*t^2 (8-18)

Now we make a slight reduction to this solution. Let us suppose that

we have a solution x of (8-16), where x has the factor t and h at the

same time has the factor t^2 (this is the case in our example since

x = 15 is divisible by three and h = 216 is divisible by 9). Then we

can write:

x = x_1 * t, h = h_1 * t^2

From (8-16), we obtain:

[(x_1)^2 * t^2] + (h_1 * t^2) = a^2

and

[(x_1)^2 * t^2] - (h_1 * t^2) = b^2

Then a^2 and b^2 are both divisible by t^2 so a and b are both

divisible by t:

a = a_1 * t, b = b_1 * t

This gives:

[(x_1)^2 * t^2] + (h_1 * t^2) = [(a_1)^2 * t^2]

and

[(x_1)^2 * t^2] - (h_1 * t^2) = [(b_1)^2 * t^2].

Cancelling the factor of t^2 gives:

(x_1)^2 + h_1 = (a_1)^2

and

(x_1)^2 - h_1 = (b_1)^2

When no further reduction is possible, we shall say that we have a

primitive solution. When this reduction is applied to (8-18), we

obtain:

x = m^2 + n^2

h = 4*m*n*(m^2 - n^2)

The numbers m and n now produce a primitive Pythagorean triple.

Thus your solution of

15^2 + 216 = 21^2

15^2 - 216 = 3^2

Is really just a multiple of the primitive solution:

5^2 + 24 = 7^2

5^2 - 24 = 1^2

This corresponds to the first row in the table here:

Congruum Problem - MathWorld - Eric Weisstein

http://mathworld.wolfram.com/CongruumProblem.html

i.e., take m = 2, n = 1. Then x = 5, h = 24, a = 7, b = 1

and as with Pythagorean triples, letting t be a natural number gives

an infinite number of solutions.

Neat problem!

I hope this is clear - let me know if I'm skipping steps where I need

to be filling in the details.

Finally, notice that writing

h = 4*m*n*(m-n)*(m+n)

is only possible when h is an element of a primitive congruum. That's

why your attempts failed. You needed to write

h = 4*m*n*(m-n)*(m+n)*t^2

which works easily when h = 216, m = 2, n = 1, t = 3.

- Doctor Paul, The Math Forum

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- Anonymous1 decade ago
2 + 2 = ( 4 ) +1 = 5

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- PuzzlingLv 71 decade ago
I'll give you a proof:

Start with the identity:

-20 = -20

Express both sides in slightly different, yet equivalent ways:

16 - 36 = 25 - 45

Factor both sides:

4^2 - 4 x 9 = 5^2 - 5 x 9

Complete the square by adding 81/4 to both sides:

4^2 - 4 x 9 + 81/4 = 5^2 - 5 x 9 + 81/4

Factor both sides again:

( 4 - 9/2 )^2 = ( 5 - 9/2 )^2

Take the square root of both sides:

4 - 9/2 = 5 - 9/2

Cancel the common factor:

4 = 5

Given:

2 + 2 = 4

And since 2 + 2 = 4 and 4 = 5, then by the transitive property:

2 + 2 = 5

Q.E.D. (But obviously there is an error in my proof... can you find it?)

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- Anonymous1 decade ago
That is from the book 1984 by George Orwell. They say that people who are brainwashed will believe this to be true, even if they know otherwise they will eventually realize that 2+2 = 5. It doesn't actually equal 5, but the power of the mind plays into believeing it.

It's a great book, read it.

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- redsox_wsLv 41 decade ago
No, thats actually from the book 1984 by Geroge Orwell, it was used to show how people were brainwashed so that if they believe 2+2 does equal 5 than that is all the proof they need.

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- Anonymous1 decade ago
Read the book 1984 by George Orwell and you will see that 2+2=5 is indeed right.

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- Anonymous1 decade ago
Its a question of freedom, how can the world exist if people don't have the power to question, and forget all logic in order to question the fact that 2+2=4. If you really get into it you will see that if you are not able to question that fact than you are not free. so in essence I am declaring that 2+2=5

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- 1 decade ago
No, its a joke. 2+2=5 for extreamly large values of 2.

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- thale138Lv 51 decade ago
its a right answer, or a possible answer, if certain things are specified. I think my teacher did something like that in algebra class. or do you mean the so called new math? You should probably work a little harder on your English skills to, from the looks of your question spelling and grammar need help.

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